LeetCode-in-Java
3628. Maximum Number of Subsequences After One Inserting
Medium
You are given a string s consisting of uppercase English letters.
You are allowed to insert at most one uppercase English letter at any position (including the beginning or end) of the string.
Return the maximum number of "LCT" subsequences that can be formed in the resulting string after at most one insertion.
Example 1:
Input: s = “LMCT”
Output: 2
Explanation:
We can insert a "L" at the beginning of the string s to make "LLMCT", which has 2 subsequences, at indices [0, 3, 4] and [1, 3, 4].
Example 2:
Input: s = “LCCT”
Output: 4
Explanation:
We can insert a "L" at the beginning of the string s to make "LLCCT", which has 4 subsequences, at indices [0, 2, 4], [0, 3, 4], [1, 2, 4] and [1, 3, 4].
Example 3:
Input: s = “L”
Output: 0
Explanation:
Since it is not possible to obtain the subsequence "LCT" by inserting a single letter, the result is 0.
Constraints:
1 <= s.length <= 105sconsists of uppercase English letters.
Solution
public class Solution {
public long numOfSubsequences(String s) {
long tc = 0;
char[] chs = s.toCharArray();
for (char c : chs) {
tc += (c == 'T') ? 1 : 0;
}
long ls = 0;
long cs = 0;
long lcf = 0;
long ctf = 0;
long lct = 0;
long ocg = 0;
long tp = 0;
for (char curr : chs) {
long rt = tc - tp;
long cg = ls * rt;
ocg = (cg > ocg) ? cg : ocg;
if (curr == 'L') {
ls++;
} else {
if (curr == 'C') {
cs++;
lcf += ls;
} else {
if (curr == 'T') {
lct += lcf;
ctf += cs;
tp++;
}
}
}
}
long fcg = ls * (tc - tp);
ocg = fcg > ocg ? fcg : ocg;
long maxi = 0;
long[] bo = {lcf, ctf, ocg};
for (long op : bo) {
maxi = op > maxi ? op : maxi;
}
return lct + maxi;
}
}